# Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6506

Exercise

Given the differentiable function

$$z=\frac{y^2}{3x}+f(xy)$$

Prove the equation

$$x^2z'_x-xyz'_y+y^2=0$$

Proof

Define

$$u=xy$$

We get the function

$$z=\frac{y^2}{3x}+f(u)$$

And the internal function

$$u(x,y)=xy$$

We will use the chain rule to calculate the partial derivatives of z.

$$z'_x=\frac{y^2}{3}\cdot (-\frac{1}{x^2})+f'_u\cdot u'_x=$$

$$=-\frac{y^2}{3x^2}+f'_u\cdot y$$

$$z'_y=\frac{1}{3x}\cdot 2y+f'_u\cdot u'_y=$$

$$=\frac{2y}{3x}+f'_u\cdot x$$

We will put the partial derivatives in the left side of the equation we need to prove.

$$x^2z'_x-xyz'_y+y^2=$$

$$=x^2(-\frac{y^2}{3x^2}+yf'_u)-xy(\frac{2y}{3x}+xf'_u)+y^2=$$

$$=-\frac{y^2}{3}+x^2yf'_u-\frac{2y^2}{3}-x^2yf'_u+y^2=$$

$$=-\frac{y^2}{3}-\frac{2y^2}{3}+y^2=$$

$$=y^2(-\frac{1}{3}-\frac{2}{3}+1)=$$

$$=y^2\cdot 0$$

$$= 0$$

We got zero as required.

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!

Share with Friends